Prove the identity

[justcurious]

Hi! I was wondering if someone could help me with solving this problem, I should prove the identity below. Thanks in advance!
(a+b+c+d)^2 + (a+b-c-d)^2 + (a+c-b-d)^2+(a+d-b-c)^2=4(a^2+b^2+c^2+d^2)

 

[tkhunny]

It should be obvious to you that (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + STUFF.
Do that four times and you have 4(a^2 + b^2 + c^2 + d^2).
Your task is to prove that all the STUFF cancels out.

 

[Dr.Peterson]

One thing that may help is to think of, for example, (a+d-b-c)^2, as ((a-b)-(c-d))^2 and keep the binomials intact for a while. You'll have fewer terms to keep track of.

 

[Cubist]

Similar to Dr.Peterson's idea, you could start by just extracting a single variable, say "a"...

(a+[b+c+d])^2 + ...
= (b+c+d)^2 + a^2 + 2*a*[b+c+d] + ...

If you do this for all the squared expressions, then the extra stuff will look like: 2*a*( [b+c+d] + [b-c-d] + [c-b-d] + ... )
Then continue to extract "b", etc

 

[pka]

Hi! I was wondering if someone could help me with solving this problem, I should prove the identity below. \((a+b+c+d)^2 + (a+b-c-d)^2 + (a+c-b-d)^2+(a+d-b-c)^2=4(a^2+b^2+c^2+d^2)\)

See expansion \((a+b+c+d)^2\)
See expansion of \((a+b+c-d)^2\)
Do you see how you can use that web resource to remove the drudge of this question?